TITLE: prove that if $\lim \limits_{n \to \infty}F( a_n)=\ell$, then $\lim \limits_{x \to \infty}F( x)=\ell$
QUESTION [5 upvotes]: Let $a_n$ be a strictly increasing sequence of positive real numbers ($a_n>0$) and $\lim \limits_{n \to \infty} a_n=\infty.$ Suppose that the sequence $(a_{n+1}-a_n)_n$ is bounded. Let $F:\mathbb{R}^+\rightarrow \mathbb R$ be a differentiable function. Ssuppose further that $\lim \limits_{x \to \infty}F'(x)=0$ and $\lim \limits_{n \to \infty}F( a_n)=\ell$. (Note that $\ell$ is a real number). Prove that: $$\lim \limits_{x \to \infty}F( x)=\ell$$ I'm stuck, I've tried to use the MVT but it gets me nowhere! can anyone help me solve this problem?

REPLY [2 votes]: Suppose $a_{n+1}-a_n \le B$ for all $n$.
Let $\epsilon>0$, and choose $K$ such that for all $x \ge K$ we have
$|F'(x)| \le {1 \over 2B} \epsilon$.
Now choose $N$ such that for $n \ge N$ we have $a_n \ge K$ and $|F(a_n) -L| < {1 \over 2 } \epsilon$. Let $K' = A_N$.
Now suppose $x \ge K'$, then $x \in [a_n,a_{n+1}]$ for some $a_n \ge K$ and
so $|F(x)-F(a_n)| \le { 1\over 2B} \epsilon |x-a_n| \le { 1\over 2} \epsilon $.
Then $|F(x)-L| \le |F(x)-F(a_n)| + |F(a_n)-L| < \epsilon$.